A tangent and a normal are drawn at the point P(2, -4) on the parabola y² = 8x, which meet the directrix of the parabola at the points A and B respectively. If Q(a, b) is a point such that AQBP is a square, then 2a + b is equal to :

Solution

Maths sir · 5 steps

P sits at the end of the latus rectum, so find A and B on x = -2, then Q is the fourth corner.

1. Read the parabola. y² = 8x is y² = 4ax with a = 2. So the directrix is x = -2. P(2, -4) lies on the curve, since 16 = 8 × 2.
2. Tangent gives A. The tangent at (x₁, y₁) is yy₁ = 4(x + x₁), so -4y = 4(x + 2), that is y = -x - 2. Putting x = -2 gives y = 0, so A is (-2, 0).
3. Normal gives B. The tangent has slope -1, so the normal has slope 1: y + 4 = x - 2, that is y = x - 6. Putting x = -2 gives y = -8, so B is (-2, -8).
4. Check the square. PA = PB = √(16 + 16) = 4√2, and the tangent and normal meet at 90° at P. So in square AQBP, P and Q are opposite corners, and AB is the other diagonal.

   Q = A + B - P = (-2 - 2 - 2, 0 - 8 + 4) = (-6, -4)

5. So. a = -6 and b = -4, so 2a + b = -12 - 4 = -16, option (a).

The idea to keep. The two diagonals of a square bisect each other. Once you know three corners, the missing one is the sum of its two neighbours minus the opposite corner.

Answer from the NTA final answer key. Solution by PrepWiser.